Angular 2路由无法正常工作

Question

我似乎无法使用Angular 2中的子项进行路由.该应用程序是我的第一个Angular 2应用程序,它非常简单,顶部有一个导航栏,可以填充页面下半部分的内容.每当我点击任何导航链接时,他们都导航到同一个孩子,甚至更糟糕的是那个孩子的内容堆叠,这意味着我点击一次,孩子加载到下面,我再次点击它再次加载下面,所以我有两次,等等.

完整内容可以在这里找到: plunker

这是我的searching.routing.ts文件:

import { ModuleWithProviders } from '@angular/core';
import { Routes, RouterModule } from '@angular/router';

import { AdvancedSearchComponent } from './advanced-search/advanced-search.component';
import { GuidedSearchComponent } from './guided-search/guided-search.component';
import { QuickSearchComponent } from './quick-search/quick-search.component';

const searchesRoutes: Routes = [
    {
        children: [
            { path: 'advanced-search', component: AdvancedSearchComponent },
            { path: 'guided-search', component: GuidedSearchComponent },
            { path: 'quick-search', component: QuickSearchComponent }
        ],
        path: '',
        component: GuidedSearchComponent
    }
];

export const searchesRouting: ModuleWithProviders = RouterModule.forChild(searchesRoutes);

这是我的app.routing.ts文件:

import { ModuleWithProviders } from '@angular/core';
import { Routes, RouterModule } from '@angular/router';

const searchesRoutes: Routes = [
    { path: 'searches', loadChildren: 'app/searches/searches.module#SearchesModule' },
    { path: '', redirectTo: "/searches", pathMatch: 'full' }
];

const appRoutes: Routes = [
    ...searchesRoutes
];

export const appRoutingProviders: any[] = [
];

export const routing: ModuleWithProviders = RouterModule.forRoot(appRoutes);

所以再次,这不能正常工作,但如果我在主文件中执行所有路由,如下所示:

import { ModuleWithProviders } from '@angular/core';
import { Routes, RouterModule } from '@angular/router';

import { AdvancedSearchComponent } from './searches/advanced-search/advanced-search.component';
import { GuidedSearchComponent } from './searches/guided-search/guided-search.component';
import { QuickSearchComponent } from './searches/quick-search/quick-search.component';

const appRoutes: Routes = [
    { path: 'searches/advanced-search', component: AdvancedSearchComponent },
    { path: 'searches/guided-search', component: GuidedSearchComponent },
    { path: 'searches/quick-search', component: QuickSearchComponent },
    { path: '', redirectTo: "/searches/guided-search", pathMatch: 'full' }
];

export const appRoutingProviders: any[] = [
];

export const routing: ModuleWithProviders = RouterModule.forRoot(appRoutes);

但我想学习如何将搜索路径委托给子模块以使其更易于管理.

Answer 1

您应该在searchesRoutes子路径中包含根组件,例如:

@Component({
    selector: 'art-search',
    template: `<router-outlet></router-outlet>`
})
export class SearchComponent {}

而且你还必须指定默认路由,如:

{ path: '', redirectTo: '/guided-search', pathMatch: 'full' }

然后您的searching.routing.ts文件将是:

const searchesRoutes: Routes = [
  {
    children: [
      { path: 'advanced-search', component: AdvancedSearchComponent },
      { path: 'guided-search', component: GuidedSearchComponent },
      { path: 'quick-search', component: QuickSearchComponent },
      { path: '', redirectTo: '/guided-search', pathMatch: 'full' }
    ],
    path: '', component: SearchComponent
  } 
];

在这里查看更多详情Plunker